Two cylinders of equal masses, one made of aluminium and the other of copper, with their lateral surfaces thermally insulated, are heated to `50^(@)C` and placed on two large blocks of ice at `0^(@)C`. If both the cylinders haver the same height, find the ratio of their depths of penetration in the ice. Assume that no heat is lost to the surroundings. Given that
`S_(Al)=0.22 "cal" gm^(-1).^(@)C^(-1), rho_(Al)=2.7gm cm^(-3)`
`S_(CU)=0.1 "cal"gm^(-1).^(@)C^(-1),rho_(Cu)=8.9gm cm^(-3)`

Answer

To solve the problem, we need to calculate the heat lost by each cylinder when they are placed on the ice and then find the ratio of the depths of penetration of the two cylinders in the ice.

### Step 1: Calculate the heat lost by the aluminum cylinder.
The heat lost by the aluminum cylinder can be calculated using the formula:
\[ Q_{Al} = m \cdot S_{Al} \cdot \Delta T \]
Where:
- \( m \) = mass of the aluminum cylinder
- \( S_{Al} = 0.22 \, \text{cal/g°C} \) (specific heat of aluminum)
- \( \Delta T = 50°C - 0°C = 50°C \) (temperature change)

### Step 2: Substitute the values for the aluminum cylinder.
Since both cylinders have equal masses, let’s denote the mass as \( m \).
\[ Q_{Al} = m \cdot 0.22 \cdot 50 \]
\[ Q_{Al} = 11m \, \text{calories} \]

### Step 3: Calculate the heat lost by the copper cylinder.
Similarly, for the copper cylinder:
\[ Q_{Cu} = m \cdot S_{Cu} \cdot \Delta T \]
Where:
- \( S_{Cu} = 0.1 \, \text{cal/g°C} \) (specific heat of copper)

### Step 4: Substitute the values for the copper cylinder.
\[ Q_{Cu} = m \cdot 0.1 \cdot 50 \]
\[ Q_{Cu} = 5m \, \text{calories} \]

### Step 5: Calculate the amount of ice melted by each cylinder.
The heat required to melt ice is given by:
\[ Q = m_{ice} \cdot L_f \]
Where \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion of ice).

For the aluminum cylinder:
\[ m_{ice, Al} = \frac{Q_{Al}}{L_f} = \frac{11m}{80} \]

For the copper cylinder:
\[ m_{ice, Cu} = \frac{Q_{Cu}}{L_f} = \frac{5m}{80} \]

### Step 6: Find the ratio of the depths of penetration.
Since both cylinders have the same height and radius, the depth of penetration into the ice is directly proportional to the mass of ice melted:
\[ \text{Depth ratio} = \frac{m_{ice, Al}}{m_{ice, Cu}} = \frac{11m/80}{5m/80} = \frac{11}{5} \]

### Final Result:
The ratio of the depths of penetration in the ice is:
\[ \text{Depth ratio} = \frac{11}{5} \]